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Vedmedyk [2.9K]

2 years ago

11

A runner exerts a force of 234 N against the ground while using 1100 W of power.How long does it take them to run a distance of

30.0 m

1 answer:

11Alexandr11 [23.1K]2 years ago

8 0

<h3>Answer:</h3>

6.38 seconds

<h3>Explanation:</h3>

We are given;

The force exerted by the runner as 234 N
Power used = 1100 W
Distance as 30.0 m

We are required to determine the time taken to run the distance;

We know that power is the rate of work done

That is;

Power = work done ÷ time

But, work done = Force × distance

Therefore;

Work done = 234 N × 30 m

= 7020 Joules

Therefore, from the formula, Power = work done ÷ time

Time = Work done ÷ power

Thus;

Time = 7020 J ÷ 1100 W

= 6.38 seconds

Hence, the time taken is 6.38 seconds

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8 0

3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

2 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

2 years ago

Calculate the force it would take to accelerate a 50 kg bike at a rate of 3 m/s2.

Ede4ka [16]

ummmm it might be 300... i used a calculator

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4 0

3 years ago

At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f

Phantasy [73]

Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of Kepler's

T ² ∝ r³

T=Time period

r=semi major axis

We also know that time period T given as

$T=\dfrac{2\pi r}{v}$

v=Speed

$v=\dfrac{2\pi r}{T}$

$v\alpha \dfrac{r}{T}$

$v\alpha \dfrac{r}{T}$

$v^2\alpha \dfrac{r^2}{T^2}$

$v^2\alpha \dfrac{r^2}{r^3}$

$v^2\alpha \dfrac{1}{r}$

$v\alpha \dfrac{1}{\sqrt{r}}$

So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.

7 0

2 years ago